Find the volume?

Question

Find the volume formed by rotating the region enclosed by:
x=8y and y^3=x with y>0
about the y-axis

Answer 1

need another boundary, unless you're talking about the points at which the two graphs intersect which is when 8y=y^3. this occurs when y=0 and y=sqrt(8), or y=2.83. those are the limits of your integral. the equation for finding the volume is determined by creating and infinite number of discs, the base area which is 8y-y^3, and a height of dy. then you integrate that equation with boundaries at y=0 and y=2.83.

integral(8y-y^3)dy=4y^2-(y^4)/4
plugging in the values for y
=(4*2.83^2-(2.83^4)/4)-(4*0^2-(0^4)/4)
=(4*8-64/4)-(0-0)
=32-16
=16

Answer 2

Basic strategy:
1. find where the two eqns meet
2. find a volume element made from the two equations
3. integrate from 0 to where the two eqns meet

1. where meet: y = x/8, y = x^(1/3)
x/8 = x^(1/3)
x^3/8^3 = x
x^2 = 512
x = sqr[512] = 16sqr[2]
2. volume element: quadrant I
dy = [x/8 - x^(1/3)]dx
in three dimensions:
dV = pi x^2 dy = pi x^2 [x/8 - x^(1/3)]dx
3. Volume = INTEGRALdV
= INTEGRAL|0-->16sqr[2]|{pi x^2 [x/8 - x^(1/3)]dx

You have to do the rest

Answer 3

v= int(outer radius)^2 - (inner radius)^2)
int = integral
(y^3)^2-(8y)^2
Pi*int(y^6-13y^2 dy)
You're using the washer method
to find the limit solve for y
8y=y^3
y(y^2-8)=0
with the boundary of (0,sqrt(8)).
integrate the integral
and you get 270.8042934